Let \(x_0\in {\mathbb {R}}^d\) , \(0<r<1\)  and let \(f\in C^2({\mathbb {R}}^d)\)  be a positive function such that
\(f(x)=\left\lbrace \begin{array}{cl}|x-x_0|^2, & |x-x_0|\le \frac{r}{2}\\r^2,& |x-x_0|\ge r\end{array}\right.\) 
and for some \(c_1 >0\)
\(|f(x)|\le c_1 r^2, \ \ \left|\frac{\partial f}{\partial x_i}(x)\right|\le c_1 r \ \ \textrm { and }\ \ \left|\frac{\partial ^2 f}{\partial x_i \partial x_j}(x)\right|\le c_1.\) 
By the optional stopping theorem we get
\(\mathbb {E}_x f(X_{t\wedge \tau _{B_r(x_0)}})-f(x_0)=\mathbb {E}^x \int _0^{t\wedge \tau _{B_r(x_0)}}\mathcal {L}f(X_s)\,ds,\ \ t>0.\) 

Let \(x\in B_r(x_0)\) . We estimate \(\mathcal {L}f(x)\)  by splitting the integral in
(REF ) into three parts.
\(\int _{B_r} &(f(x+h)-f(x)-\nabla f(x)\cdot h \mathbb {1}_{\lbrace |h|\le 1\rbrace })K(x,h)\,dh\\& \le c_2\int _{B_r}|h|^2K(x,h)\,dh\le c_2\kappa \int _{B_r}|h|^{2-d}\ell (|h|)\,dh \le c_3 r^{2}\ell (r),\) 

where in the last line we have used Karamata's theorem, cf. property
(REF ) in sec:appendix. On the other hand, on \(B_r^c\)  we
have
\(\int _{B_r^c} &(f(x+h)-f(x))K(x,h)\,dh\le 2\Vert f\Vert _\infty \int _{B_r^c}K(x,h)\,dh\\& \le 2\Vert f\Vert _\infty \left(\kappa \int _{B_1\setminus B_r}|h|^{-d}\ell (|h|)\,dh+\int _{B_1^c}K(x,h)\,dh\right) \le c_4r^{2}L(r)\,dr\,,\) 

where we applied property (REF ) from sec:appendix.
Last, we estimate
\(\left|\int _{B_1\setminus B_r}h\cdot \nabla f(x) K(x,h)\,dh\right|&\le c_1 r\int _{B_1\setminus B_r} |h| K(x,h)\,dh\\&\le c_1 \kappa r \int _{B_1\setminus B_r} |h|^{-d+1}\ell (|h|)\,dh \le c_5 r^2\ell (r),\) 

by Karamata's theorem again. Therefore, by property (REF ) from
sec:appendix we conclude that
there is a constant \(c_6>0\)  such that for all \(x\in B_r(x_0)\)  and \(r\in (0,1)\) 
we have
\(\mathcal {L}f(x)\le c_6 r^2 L(r).\) 
Let us look again at (REF ). On
\(\lbrace \tau _{B_r(x_0)}\le t\rbrace \)  we have \(X_{t\wedge \tau _{B_r(x_0)}}\in B_r(x_0)^c\)  and so
\(f(X_{t\wedge \tau _{B_r(x_0)}})\ge r^2\) . Thus, by (REF ) and
(REF ) we get
\(\mathbb {P}_{x_0}(\tau _{B_r(x_0)}\le t)\le c_6 L(r)t.\) 
\(\Box \)
There are constants \(C_2>0\)  and \(C_3>0\)  such that for \(x_0\in {\mathbb {R}}^d\)
\(\sup _{x\in {\mathbb {R}}^d}\mathbb {E}_x\tau _{B_r(x_0)}\le \frac{C_2}{L(r)}\,,\quad r\in (0,1/2)\) 
and
\(\inf _{x\in B_{r/2}(x_0)}\mathbb {E}_x\tau _{B_r(x_0)}\ge \frac{C_3}{L(r)}\,,\quad r\in (0,1)\) 
The proof is similar to the proof of the exit time estimates in [1]}.
(a) First we prove the upper estimate for the exit time.
Let \(x\in {\mathbb {R}}^d\) , \(r\in (0,1/2)\)  and let
\(S=\inf \lbrace t>0|\, |X_t-X_{t-}|>2r\rbrace \) 
be the first time of a jump larger than \(2r\) . With the help of the
Lévy system formula (cf. [1]}) and () we
can deduce
\(\mathbb {P}_x(S\le L(r)^{-1}) & =\mathbb {E}_x\sum _{t\le L(r)^{-1}\wedge S}\mathbb {1}_{\lbrace |X_t-X_{t-}|>2r\rbrace } =\mathbb {E}_x\int \limits _0^{L(r)^{-1}\wedge S}\int \limits _{B_{2r}^c}K(X_s,h)\,dh\,ds \nonumber \\&\ge c_1\mathbb {E}_x[L(r)^{-1}\wedge S]\int \limits _{2r}^1\frac{\ell (t)}{t}\,dt\,.\) 


Since \(L\)  is regularly varying at zero,
\(\mathbb {E}_x[L(r)^{-1}\wedge S]& \ge L(r)^{-1}\mathbb {P}_x(S>L(r)^{-1}) \ge c_2L(2r)^{-1}\big (1-\mathbb {P}_x(S\le L(r)^{-1})\big )\,\) 

and so it follows from (REF ) that
\(\mathbb {P}_x(S\le L(r)^{-1})\ge c_3\) 
with \(c_3=\frac{c_1c_2}{c_1c_2+1} \in (0,1)\) . The strong Markov
property and (REF ) lead to
\(\mathbb {P}_x(S>m L(r)^{-1})\le (1-c_3)^m,\ \ m\in {\mathbb {N}}\,.\) 
Since \(\tau _{B_r(x_0)}\le S\) ,
\(\mathbb {E}_x\tau _{B_r(x_0)}\le \mathbb {E}_x S &\le L(r)^{-1}\sum _{m=0}^\infty (m+1)\mathbb {P}_x(S>L(r)^{-1}m)\\&\le L(r)^{-1}\sum _{m=0}^\infty (m+1)(1-c_3)^m\,.\) 
(b) Now we prove the lower estimate of the exit time. Let \(r\in (0,1)\)  and
\(y\in B_{r/2}(x_0)\) .
By prop:estp,
\(\mathbb {P}_y(\tau _{B_r(x_0)}\le t)\le \mathbb {P}_y(\tau _{B_{r/2}(y)}\le t)\le C_1tL(r/2),\quad t>0\,,\) 
since \(B_{r/2}(y)\subset B_r(x_0)\)  . Choose \(t=\frac{1}{2C_1L(r/2)}\) . Then
\(\mathbb {E}_y\tau _{B_r(x_0)}&\ge \mathbb {E}_y[\tau _{B_r(x_0)};\tau _{B_r(x_0)}>t]\ge t\mathbb {P}_y(\tau _{B_r(x_0)}>t)\\&\ge t(1-C_1L(r/2)t)=\frac{1}{4C_1L(r/2)}\,.\) 

By (REF ) from sec:appendix we know that
\(L\)  is regularly varying at zero. Hence there is a
constant \(c_1>0\)  such that \(L(r/2)\le c_1 L(r)\)  for all \(r\in (0,1/2)\) .
Therefore
\(\mathbb {E}_y\tau _{B_r(x_0)}\ge \frac{1}{4C_1c_1L(r)}\)  .
\(\Box \)
There is a constant \(C_4>0\)  such that for all \(x_0\in {\mathbb {R}}^d\)  and \(r,s\in (0,1)\)  satisfying \(2r<s\)
\(\sup _{x\in B_r(x_0)}\mathbb {P}_x(X_{\tau _{B_r(x_0)}}\notin B_s(x_0))\le C_4 \frac{L(s)}{L(r)}\,.\) 
Let \(x_0 \in {\mathbb {R}}^d\) , \(r,s \in (0,1)\)  and \(x\in B_r(x_0)\) . Set
\(B_r:=B_r(x_0)\) . By the Lévy system formula, for \(t>0\)
\(\mathbb {P}_x(X_{\tau _{B_r}\wedge t}\notin B_s)&=\mathbb {E}_x\sum \limits _{s\le \tau _{B_r}\wedge t} \mathbb {1}_{\lbrace X_{s-}\in B_r,X_s\in B_s^c\rbrace } =\mathbb {E}_x\int \limits _0^{\tau _{B_r}\wedge t}\int \limits _{B_s^c} K(X_s,z-X_s)\,dz\,ds \,.\) 
Let \(y\in B_r\) . Since \(s\ge 2r\) , it follows that
\(B_{s/2}(y)\subset B_s\)  and hence
\(\int \limits _{B_s^c} K(y,z-y)\,dz&\le \int \limits _{B_{s/2}(y)^c} K(y,z-y)\,dz \le c_1\int _{s/2}^1 \frac{\ell (u)}{u}\,du+c_2 \le c_3L(s)\,.\) 

where in the last inequality we have used that \(L\)  varies
regularly at zero and that \(\lim \limits _{r\rightarrow 0+}L(r)> 0\) , cf.
(REF ) in sec:appendix.
The above considerations together with prop:prob-2 imply
\(\mathbb {P}_x(X_{\tau _{B_r}\wedge t}\notin B_s)\le c_3 L(s)\mathbb {E}_x \tau _{B_r}\le c_4 \frac{L(s)}{L(r)}\,.\) 
Letting \(t\rightarrow \infty \)  we obtain the desired estimate.
\(\Box \)
For \(x_0\in {\mathbb {R}}^d\)  and \(r\in (0,1)\)  we define the following measure
\(\mu _{x_0,r}(dx)=\frac{\ell (|x-x_0|)}{L(|x-x_0|)}\,|x-x_0|^{-d}\mathbb {1}_{\lbrace r\le |x-x_0|<1\rbrace }\,dx\,.\) 
Define \(\varphi _a(r)=L^{-1}(\frac{1}{a}L(r))\)  for \(r\in (0,1)\)  and \(a>1\) . The
following property is important for the construction below:
\(r = L^{-1}(L(r)) \le L^{-1}(\tfrac{1}{a}L(r)) = \varphi _a(r)\,.\) 
Now we can prove a Krylov-Safonov type hitting estimate which
includes prop:hittingnew as a special case.
There exists a constant \(C_5>0\)  such that for all \(x_0\in {\mathbb {R}}^d\) , \(a>1\) , \(r\in (0,\frac{1}{2})\)  and
\(A\subset B_{\varphi _a(r)}(x_0)\setminus B_r(x_0)\)  satisfying
\(\mu _{x_0,r}(A)\ge \frac{1}{2}\mu _{x_0,r}(B_{\varphi _a(r)}(x_0) \setminus B_r(x_0))\)
\(\mathbb {P}_y(T_A<\tau _{B_{\varphi _a(r)}(x_0)})\ge \mathbb {P}_y(X_{\tau _{B_r(x_0)}}\in A)\ge C_5\frac{\ln {a}}{a}\,,\quad y\in B_{r/2}(x_0)\,.\) 
Consider \(x_0\in {\mathbb {R}}^d\) , \(a>1\) , \(r\in (0,\frac{1}{2})\)  and a set
\(A\subset B_{\varphi _a(r)}(x_0)\setminus B_r(x_0)\)  satisfying
\(\mu _{x_0,r}(A)\ge \frac{1}{2}\mu _{x_0,r}(B_{\varphi _a(r)}(x_0) \setminus B_r(x_0))\) . Set \(\mu :=\mu _{x_0,r}\) , \(\varphi :=\varphi _a\) , \(B_s:=B_s(x_0)\)  and
let \(y\in B_{r/2}\) . The first
inequality follows from
\(\lbrace X_{\tau _{B_r}}\in A\rbrace \subset \lbrace T_A<\tau _{B_{\varphi (r)}}\rbrace \)  since \(A\subset B_{\varphi (r)}\setminus B_r\)  .
By the Lévy system formula, for \(t>0\) ,
\(\mathbb {P}_y(X_{\tau _{B_r}\wedge t}\in A)&=\mathbb {E}_y\sum \limits _{s\le \tau _{B_r}\wedge t} \mathbb {1}_{\lbrace X_{s-}\in B_r,X_s\in A\rbrace } =\mathbb {E}_y\int \limits _0^{\tau _{B_r}\wedge t}\int \limits _{A} K(X_s,z-X_s)\,dz\,ds\,.\) 
Since \(|z-x|\le |z-x_0|+|x_0-x|\le |z-x_0|+r\le 2|z-x_0|\)  for \(x\in B_r\)  and \(z\in B_r^c\) ,
\(\mathbb {E}_y\int \limits _0^{\tau _{B_r}\wedge t}\int \limits _{A} K(X_s,z-X_s)\,dz\,ds\ge c_1\mathbb {E}_y[\tau _{B_r}\wedge t]\int _A \frac{\ell (|z-x_0|)}{|z-x_0|^d}\,dz\,,\) 
where we have used property (REF ) given in
sec:appendix.
Since \(L\)  is decreasing,
\(\int _A \frac{\ell (|z-x_0|)}{|z-x_0|^d}\,dz&=\int _AL(|z-x_0|)\mu (dz) \ge L(\varphi (r))\mu (A)\ge \frac{L(r)}{2a}\mu (B_{\varphi (r)}\setminus B_r)\,.\) 

Noting that
\(\mu (B_{\varphi (r)}\setminus B_r)=c_2\int _r^{\varphi (r)}\frac{1}{L(s)}\frac{\ell (s)\,ds}{s}=-c_2\ln L(s)|_r^{\varphi (r)}=c_2\ln a\,,\) 
we conclude from
(REF )–(REF ) that
\(\mathbb {P}_y(T_A<\tau _{B_{\varphi _a(r)}(x_0)})\ge c_3L(r)\frac{\ln {a}}{a}\mathbb {E}_y[\tau _{B_r}\wedge t]\,.\) 
Letting \(t\rightarrow \infty \)  and using the lower bound in prop:prob-2 we get
\(\mathbb {P}_y(T_A<\tau _{B_{\varphi _a(r)}(x_0)}) \ge c_3L(r)\,\frac{\ln {a}}{a}\,\mathbb {E}_y\tau _{B_r} \ge c_3L(r)\,\frac{\ln {a}}{a}\,C_3L(r)^{-1}=c_3C_3\frac{\ln {a}}{a}\,.\) 
\(\Box \)
Reglarity of harmonic functions
Let \(x_0 \in {\mathbb {R}}^d\) , \(r\in (0,\frac{1}{2})\) , \(x\in B_{r/4}(x_0)\) . Using
(REF ) from sec:appendix with \(\delta =1\) , we see
that there is a constant
\(c_0\ge 1\)  so that
\(\frac{L(s)}{L(s^{\prime })}\le c_0\left(\frac{s}{s^{\prime }}\right)^{-\alpha -1},\quad 0<s<s^{\prime }<1\,.\) 
Define for \(n\in {\mathbb {N}}\)
\(r_n:=L^{-1}( L(\tfrac{r}{2})a^{n-1})\quad \text{ and }\quad s_n:=3 \Vert u\Vert _\infty b^{-(n-1)}\) 
for some constants \(b\in (1,\frac{3}{2})\)  and \(a>c_0 2^{\alpha +1}\)  that will be
chosen in the proof
independently of \(n\) , \(r\)  and \(u\) . As we explained in the
introduction, thm:main trivially holds true of \(\lim \limits _{r\rightarrow 0+}L(r)\) 
is finite. Thus, we can assume \(\lim \limits _{r\rightarrow 0+} L(r)\)  to be infinite. This
implies that \(r_n \rightarrow 0\)  for \(n \rightarrow \infty \)  as it should be.
We will use the following abbreviations:
\(B_n:=B_{r_n}(x), \quad \tau _n:=\tau _{B_n}, \quad m_n:=\inf _{B_n}u,\quad M_n:=\sup _{B_n}u\,.\) 
We are going to prove
\(M_k-m_k\le s_k\) 
for all \(k\ge 1\) .
Assume for a moment that (REF ) is proved. Then, for any \(r\in (0,\frac{1}{2})\)  and \(y\in B_{r/4}(x_0)\subset B_{r/2}(x)\)  we can find \(n\in {\mathbb {N}}\)  so that
\(r_{n+1}\le |y-x|< r_n\,.\) 
Furthermore, since \(L\)  is decreasing, we obtain with
\(\gamma =\frac{\ln {b}}{\ln {a}}\in (0,1)\)
\(|u(y)-u(x)|&\le s_n=3 b\Vert u\Vert _\infty a^{-n\frac{\ln b}{\ln a}} =3b\Vert u\Vert _\infty \left[\frac{L(r_{n+1})}{L(\frac{r}{2})}\right]^{-\frac{\ln {b}}{\ln {a}}} \le 3b\Vert u\Vert _\infty \left[\frac{L(|x-y|)}{L(\frac{r}{2})}\right]^{-\gamma }\,,\) 

which proves our assertion. Thus it remains to prove
(REF ).
We are going to prove (REF ) by an inductive argument.
Obviously, \(M_1-m_1\le 2\Vert u\Vert _\infty \le s_1\) . Since \(1<b<\frac{3}{2}\) , it follows that
\(M_2-m_2\le 2\Vert u\Vert _\infty \le 3\Vert u\Vert _\infty b^{-1}=s_2\,.\) 
Assume now that (REF ) is true for all \(k\in \lbrace 1,2,\ldots ,n\rbrace \)  for some \(n\ge 2\) .
Let \(\varepsilon >0\)  and take \({z_1},z_2 \in B_{n+1}\)  so that
\(u({z_1})\le m_{n+1}+\frac{\varepsilon }{2}\ \ \ \ \ \ \ \ u(z_2 )\ge M_{n+1}-\frac{\varepsilon }{2}\,.\) 
It is enough to show that
\(u(z_2 )-u({z_1})\le s_{n+1},\) 
since then
\(M_{n+1}-m_{m+1}-\varepsilon \le s_{n+1},\) 
which implies (REF ) for \(k=n+1\) , since \(\varepsilon >0\)  was arbitrary.
By the optional stopping theorem,
\(u(z_2 )-u({z_1})=&\ \mathbb {E}_{z_2}[u(X_{\tau _n})-u({z_1})]\nonumber \\=&\ \mathbb {E}_{z_2}[u(X_{\tau _n})-u({z_1});X_{\tau _n}\in B_{n-1}]\\&+\sum \limits _{i=1}^{n-2}\mathbb {E}_{z_2}[u(X_{\tau _n})-u({z_1});X_{\tau _n}\in B_{n-i-1}\setminus B_{n-i}]\\&+\mathbb {E}_{z_2}[u(X_{\tau _n})-u({z_1});X_{\tau _n}\in B_1^c]=I_1+I_2+I_3\,.\) 
Let \(A=\lbrace z \in B_{n-1}\setminus B_n|\, u(z)\le \frac{m_n+M_n}{2}\rbrace \) . It is
sufficient to consider the case \(\mu _{x,r_n}(A)\ge \frac{1}{2}\mu _{x,r_n}(B_{n-1}\setminus B_n)\) , where
\(\mu _{x,r}\)  is the measure defined by (REF ). In the
remaining case we would use
\(\mu _{x,r_n}((B_{n-1}\setminus B_n)\setminus A)\ge \frac{1}{2}\mu _{x,r_n}(B_{n-1}\setminus B_n)\) 
and could continue the proof with \(\Vert u\Vert _\infty -u\)  and
\((B_{n-1}\setminus B_n)\setminus A=\left\lbrace z\in B_{n-1}\setminus B_n|\, \Vert u\Vert _\infty -u(z)\le \frac{\Vert u\Vert _\infty -m_n+\Vert u\Vert _\infty -M_n}{2}\right\rbrace \) 
instead of \(u\)  and \(A\) .
The estimate (REF ) implies
\(a=\tfrac{L(r_{n+1})}{L(r_n)}\le c_0(\tfrac{r_{n+1}}{r_n})^{-\alpha -1}\) , from
where we deduce \(r_{n+1}\le r_n (c_0a^{-1})^{\frac{1}{\alpha +1}} \le \frac{r_n}{2}\)  because of \(a>c_0 2^{\alpha +1}\) . Next, we make use of the
following property:
\(r_{n-1} = L^{-1}(L(\tfrac{r}{2}) a^{n-2}) = L^{-1}(\tfrac{1}{a}L(\tfrac{r}{2}) a^{n-1}) = L^{-1}(\tfrac{1}{a} L(r_n)) = \varphi _a(r_n) \,.\) 

Then by
prop:hittinggeneral (with
\(r=r_n\)  and \(x_0=x\) ) we get
\(p_n:=\mathbb {P}_{z_2}(X_{\tau _n}\in A)\ge C_5\frac{\ln {a}}{a}\,.\) 
Hence,
\(I_1&=\mathbb {E}_{z_2}[u(X_{\tau _n})-u({z_1});X_{\tau _n}\in B_{n-1}]\\&=\mathbb {E}_{z_2}[u(X_{\tau _n})-u({z_1});X_{\tau _n}\in A]+\mathbb {E}_{z_2}[u(X_{\tau _n})-u({z_1});X_{\tau _n}\in B_{n-1}\setminus A]\\&\le \left(\tfrac{m_n+M_n}{2}-m_n\right)p_n+s_{n-1}(1-p_n)\\&\le \tfrac{1}{2}s_n p_n+s_{n-1}(1-p_n)\le s_{n-1}(1-\tfrac{1}{2}p_n)\le s_{n-1}(1-\tfrac{C_5\ln {a}}{2a})\,.\) 
By prop:prob-3,
\(I_2& \le \sum \limits _{i=1}^{n-2} s_{n-i-1}\mathbb {P}_{z_2}(X_{\tau _n}\notin B_{n-i})\le C_4 \sum \limits _{i=1}^{n-2} s_{n-i-1}\tfrac{L(r_{n-i})}{L(r_n)}\\&\le 3C_4\Vert u\Vert _\infty \sum \limits _{i=1}^{n-2}b^{-(n-i-2)}\tfrac{a^{n-i-1}}{a^{n-1}}\le 3C_4 \Vert u\Vert _\infty \tfrac{b^{-n+3}}{a-b}\\&\le C_4 \tfrac{b^3}{a-b}s_{n+1}\,.\) 

Similarly, by prop:prob-3,
\(I_3\le 2\Vert u\Vert _\infty \mathbb {P}_{z_2}(X_{\tau _n}\notin B_1)\le 2C_4\Vert u\Vert _\infty \tfrac{L(r_1)}{L(r_n)}=\tfrac{2C_4 }{3}b\left(\tfrac{b}{a}\right)^{n-1}s_{n+1}\le C_4\tfrac{b^2}{a} s_{n+1}\,.\) 
Hence,
\(u(z_2)-u(z_1)\le s_{n+1}b^2\left[1-\tfrac{C_5\ln {a}}{2a}+ \tfrac{C_4 b}{a-b}+\tfrac{C_4 }{a}\right]\,.\) 
Since \(a-b\ge \frac{a}{4}\)  for \(b\in (1,\frac{3}{2})\)  and
\(a>c_0 2^{\alpha +1}\ge 2\) , it follows
that
\(q:=1-\tfrac{C_5\ln {a}}{2a}+\tfrac{C_4 b}{a-b}+\tfrac{C_4 }{a}\le 1-\tfrac{C_5\ln {a}}{2a}+\tfrac{7C_4 }{a}=1-\tfrac{C_5\ln a-14C_4 }{2a}\,.\) 
Next, we choose \(a>c_0 2^{\alpha +1}\)  so large that \(C_5\ln {a}-14C_4 >0\) . Thus
\(q<1\) .
Finally, we choose \(b\in (1,\frac{3}{2})\)  sufficiently small so that \(b^2q<1\)  .
Hence, (REF ) holds, which finishes the proof of the inductive step and the theorem .
\(\Box \)
appendix

Slow and Regular Variation
In this section we collect some properties of slowly resp. regularly varying
functions that are used in our main arguments. Moreover we list several examples
which illustrate the range of application of our approach.
A measurable and positive function \(\ell \colon (0,1)\rightarrow (0,\infty )\)  is
said to vary regularly at zero with index \(\rho \in {\mathbb {R}}\)  if for every \(\lambda >0\)
\(\lim _{r\rightarrow 0+}\frac{\ell (\lambda r)}{\ell (r)}=\lambda ^{\rho }\,.\) 
If a function varies regularly at zero with index 0 it is said to vary
slowly at zero. For simplicity, we call such functions regularly
varying resp. slowly varying functions.
Note that slowly resp. regularly varying functions include functions which are
neither
increasing nor decreasing. By [3]} it follows that
any function \(\ell \)  that
varies regularly with index \(\rho \in {\mathbb {R}}\)  is of the form
\(\ell (r)=r^{\rho }\ell _0(r)\)  for some
function \(\ell _0\)  that varies slowly.
Assume \(\int _0^1 s \, \ell (s) \, \, \textnormal {d}s \le c\)  for some \(c >0\) . Let \(L\colon (0,1)\rightarrow (0,\infty )\)  be defined by
\(L(r)=\int \limits _r^1\frac{\ell (s)}{s}\, \, \textnormal {d}s\ .\) 
The function \(L\)  is well defined because \(L(r) = r^{-2} \int _r^1r^2 \frac{\ell (s)}{s}\, \, \textnormal {d}s \le r^{-2} \int _r^1s \ell (s) \, \, \textnormal {d}s \le c r^{-2}\) . Note that (REF ) and
() imply that \(\int _0^1 s \, \ell (s) \, \, \textnormal {d}s \le c\)  does hold in our
setting. We note that the function \(L\)  is always decreasing.
Let us list further properties which are making use of in our proofs. Note that
they are established [3]} for functions which are slowly resp. regularly
varying at the point \(+\infty \) . By a simple inversion we adopt the results
to functions which are slowly resp. regularly
varying at the point 0.

 If \(\ell \)  is slowly varying, then
[3]} \(L\)  is slowly varying with
\(\lim \limits _{r\rightarrow 0+}L(r)=+\infty \qquad \text{ and }\qquad \lim \limits _{r\rightarrow 0+}\frac{\ell (r)}{L(r)}=0\,.\) 

 If \(\ell \)  is slowly varying and \(\rho >-1\) , then
Karamata's theorem [3]} ensures
\(\lim _{r\rightarrow 0+}\frac{\int _0^rs^{\rho }\ell (s)\,ds}{r^{\rho +1}\ell (r)}=(\rho +1)^{-1}\,.\) 

 If \(\ell \)  is
regularly varying of order \(-\alpha < 0\)  (in our case \(0 < \alpha < 2\) ),
then
[3]}
\(\lim _{r\rightarrow 0+}\frac{L(r)}{\ell (r)}=\alpha ^{-1}\,.\) 
In particular, if \(\ell \)  is regularly varying of order \(-\alpha < 0\) ,
then so is \(L\) .

 Assume \(\ell \)  is regularly varying of order
\(-\alpha \le 0\)  and stays bounded away from 0 and \(+\infty \)  on every compact
subset of \((0,1)\) . Then Potter's theorem [3]} implies
that for every \(\delta >0\)  there is a
constant
\(C=C(\delta )\ge 1\)  such that for \(r,s \in (0,1)\)
\(\frac{\ell (r)}{\ell (s)}\le C\max \left\lbrace  \left(\frac{r}{s}\right)^{-\alpha -\delta },\left(\frac{r}{s}\right)^{-\alpha +\delta } \right\rbrace  \,.\) 

 Since \(L\)  is nonincreasing, we observe
\(\lim \limits _{r\rightarrow 0+} L(r) \in (0,+\infty ]\) .

<TABLE>Let us look at different choices for the function \(\ell \) , given in Table
REF . Here \(\beta \in (0,2)\) , \(a >1\)  are fixed. We list six examples of a function \(s \mapsto \ell _i(s)\)  together
with \(s\mapsto L_i(s)\)  and \(s \mapsto \varphi _a(s)=L_i^{-1}(\frac{1}{a}L_i(s))\) . Recall that the function \(\varphi _a\)  appears in
prop:hittingnew and determines the scaling that we are using, see
also property (REF ) and the definition of \(r_n\)  in the proof of
thm:main. Note that case No. 6 is significantly different from the
other
cases. Both, the integral \(\int _{B_1}|h|^{-d} \ell _6(|h|) \, \, \textnormal {d}h\)  and the
expression \(\lim \limits _{s\rightarrow 0+} L_6 (s)\)  are finite. Moreover, the limit
\(\lim \limits _{s\rightarrow 0+} L_6^{-1}(\frac{1}{a} L_6(s))\)  is not equal to zero. These differences
reflect the
fact that the corresponding operator in (REF ) has an integrable
kernel. Recall that
prop:relation relates the behavior of the function \(L\)  close
to the origin to the behaviour of the multiplier of the operator (in the case
of constant coefficents) for large values of \(|\xi |\) . In the case No. 6 the
multiplier stays bounded.
Acknowledgements: We thank T. Grzywny for a helpful comment on the limit
case \(\alpha =2\) .
